Permutations#
- class sympy.combinatorics.permutations.Permutation(*args, size=None, **kwargs)[source]#
A permutation, alternatively known as an ‘arrangement number’ or ‘ordering’ is an arrangement of the elements of an ordered list into a one-to-one mapping with itself. The permutation of a given arrangement is given by indicating the positions of the elements after re-arrangement [R80]. For example, if one started with elements
[x, y, a, b]
(in that order) and they were reordered as[x, y, b, a]
then the permutation would be[0, 1, 3, 2]
. Notice that (in SymPy) the first element is always referred to as 0 and the permutation uses the indices of the elements in the original ordering, not the elements(a, b, ...)
themselves.>>> from sympy.combinatorics import Permutation >>> from sympy import init_printing >>> init_printing(perm_cyclic=False, pretty_print=False)
Permutations Notation
Permutations are commonly represented in disjoint cycle or array forms.
Array Notation And 2-line Form
In the 2-line form, the elements and their final positions are shown as a matrix with 2 rows:
[0 1 2 … n-1] [p(0) p(1) p(2) … p(n-1)]
Since the first line is always
range(n)
, where n is the size of p, it is sufficient to represent the permutation by the second line, referred to as the “array form” of the permutation. This is entered in brackets as the argument to the Permutation class:>>> p = Permutation([0, 2, 1]); p Permutation([0, 2, 1])
Given i in range(p.size), the permutation maps i to i^p
>>> [i^p for i in range(p.size)] [0, 2, 1]
The composite of two permutations p*q means first apply p, then q, so i^(p*q) = (i^p)^q which is i^p^q according to Python precedence rules:
>>> q = Permutation([2, 1, 0]) >>> [i^p^q for i in range(3)] [2, 0, 1] >>> [i^(p*q) for i in range(3)] [2, 0, 1]
One can use also the notation p(i) = i^p, but then the composition rule is (p*q)(i) = q(p(i)), not p(q(i)):
>>> [(p*q)(i) for i in range(p.size)] [2, 0, 1] >>> [q(p(i)) for i in range(p.size)] [2, 0, 1] >>> [p(q(i)) for i in range(p.size)] [1, 2, 0]
Disjoint Cycle Notation
In disjoint cycle notation, only the elements that have shifted are indicated.
For example, [1, 3, 2, 0] can be represented as (0, 1, 3)(2). This can be understood from the 2 line format of the given permutation. In the 2-line form, [0 1 2 3] [1 3 2 0]
The element in the 0th position is 1, so 0 -> 1. The element in the 1st position is three, so 1 -> 3. And the element in the third position is again 0, so 3 -> 0. Thus, 0 -> 1 -> 3 -> 0, and 2 -> 2. Thus, this can be represented as 2 cycles: (0, 1, 3)(2). In common notation, singular cycles are not explicitly written as they can be inferred implicitly.
Only the relative ordering of elements in a cycle matter:
>>> Permutation(1,2,3) == Permutation(2,3,1) == Permutation(3,1,2) True
The disjoint cycle notation is convenient when representing permutations that have several cycles in them:
>>> Permutation(1, 2)(3, 5) == Permutation([[1, 2], [3, 5]]) True
It also provides some economy in entry when computing products of permutations that are written in disjoint cycle notation:
>>> Permutation(1, 2)(1, 3)(2, 3) Permutation([0, 3, 2, 1]) >>> _ == Permutation([[1, 2]])*Permutation([[1, 3]])*Permutation([[2, 3]]) True
Caution: when the cycles have common elements between them then the order in which the permutations are applied matters. This module applies the permutations from left to right.
>>> Permutation(1, 2)(2, 3) == Permutation([(1, 2), (2, 3)]) True >>> Permutation(1, 2)(2, 3).list() [0, 3, 1, 2]
In the above case, (1,2) is computed before (2,3). As 0 -> 0, 0 -> 0, element in position 0 is 0. As 1 -> 2, 2 -> 3, element in position 1 is 3. As 2 -> 1, 1 -> 1, element in position 2 is 1. As 3 -> 3, 3 -> 2, element in position 3 is 2.
If the first and second elements had been swapped first, followed by the swapping of the second and third, the result would have been [0, 2, 3, 1]. If, you want to apply the cycles in the conventional right to left order, call the function with arguments in reverse order as demonstrated below:
>>> Permutation([(1, 2), (2, 3)][::-1]).list() [0, 2, 3, 1]
Entering a singleton in a permutation is a way to indicate the size of the permutation. The
size
keyword can also be used.Array-form entry:
>>> Permutation([[1, 2], [9]]) Permutation([0, 2, 1], size=10) >>> Permutation([[1, 2]], size=10) Permutation([0, 2, 1], size=10)
Cyclic-form entry:
>>> Permutation(1, 2, size=10) Permutation([0, 2, 1], size=10) >>> Permutation(9)(1, 2) Permutation([0, 2, 1], size=10)
Caution: no singleton containing an element larger than the largest in any previous cycle can be entered. This is an important difference in how Permutation and Cycle handle the
__call__
syntax. A singleton argument at the start of a Permutation performs instantiation of the Permutation and is permitted:>>> Permutation(5) Permutation([], size=6)
A singleton entered after instantiation is a call to the permutation – a function call – and if the argument is out of range it will trigger an error. For this reason, it is better to start the cycle with the singleton:
The following fails because there is no element 3:
>>> Permutation(1, 2)(3) Traceback (most recent call last): ... IndexError: list index out of range
This is ok: only the call to an out of range singleton is prohibited; otherwise the permutation autosizes:
>>> Permutation(3)(1, 2) Permutation([0, 2, 1, 3]) >>> Permutation(1, 2)(3, 4) == Permutation(3, 4)(1, 2) True
Equality Testing
The array forms must be the same in order for permutations to be equal:
>>> Permutation([1, 0, 2, 3]) == Permutation([1, 0]) False
Identity Permutation
The identity permutation is a permutation in which no element is out of place. It can be entered in a variety of ways. All the following create an identity permutation of size 4:
>>> I = Permutation([0, 1, 2, 3]) >>> all(p == I for p in [ ... Permutation(3), ... Permutation(range(4)), ... Permutation([], size=4), ... Permutation(size=4)]) True
Watch out for entering the range inside a set of brackets (which is cycle notation):
>>> I == Permutation([range(4)]) False
Permutation Printing
There are a few things to note about how Permutations are printed.
Deprecated since version 1.6: Configuring Permutation printing by setting
Permutation.print_cyclic
is deprecated. Users should use theperm_cyclic
flag to the printers, as described below.1) If you prefer one form (array or cycle) over another, you can set
init_printing
with theperm_cyclic
flag.>>> from sympy import init_printing >>> p = Permutation(1, 2)(4, 5)(3, 4) >>> p Permutation([0, 2, 1, 4, 5, 3])
>>> init_printing(perm_cyclic=True, pretty_print=False) >>> p (1 2)(3 4 5)
2) Regardless of the setting, a list of elements in the array for cyclic form can be obtained and either of those can be copied and supplied as the argument to Permutation:
>>> p.array_form [0, 2, 1, 4, 5, 3] >>> p.cyclic_form [[1, 2], [3, 4, 5]] >>> Permutation(_) == p True
3) Printing is economical in that as little as possible is printed while retaining all information about the size of the permutation:
>>> init_printing(perm_cyclic=False, pretty_print=False) >>> Permutation([1, 0, 2, 3]) Permutation([1, 0, 2, 3]) >>> Permutation([1, 0, 2, 3], size=20) Permutation([1, 0], size=20) >>> Permutation([1, 0, 2, 4, 3, 5, 6], size=20) Permutation([1, 0, 2, 4, 3], size=20)
>>> p = Permutation([1, 0, 2, 3]) >>> init_printing(perm_cyclic=True, pretty_print=False) >>> p (3)(0 1) >>> init_printing(perm_cyclic=False, pretty_print=False)
The 2 was not printed but it is still there as can be seen with the array_form and size methods:
>>> p.array_form [1, 0, 2, 3] >>> p.size 4
Short Introduction To Other Methods
The permutation can act as a bijective function, telling what element is located at a given position
>>> q = Permutation([5, 2, 3, 4, 1, 0]) >>> q.array_form[1] # the hard way 2 >>> q(1) # the easy way 2 >>> {i: q(i) for i in range(q.size)} # showing the bijection {0: 5, 1: 2, 2: 3, 3: 4, 4: 1, 5: 0}
The full cyclic form (including singletons) can be obtained:
>>> p.full_cyclic_form [[0, 1], [2], [3]]
Any permutation can be factored into transpositions of pairs of elements:
>>> Permutation([[1, 2], [3, 4, 5]]).transpositions() [(1, 2), (3, 5), (3, 4)] >>> Permutation.rmul(*[Permutation([ti], size=6) for ti in _]).cyclic_form [[1, 2], [3, 4, 5]]
The number of permutations on a set of n elements is given by n! and is called the cardinality.
>>> p.size 4 >>> p.cardinality 24
A given permutation has a rank among all the possible permutations of the same elements, but what that rank is depends on how the permutations are enumerated. (There are a number of different methods of doing so.) The lexicographic rank is given by the rank method and this rank is used to increment a permutation with addition/subtraction:
>>> p.rank() 6 >>> p + 1 Permutation([1, 0, 3, 2]) >>> p.next_lex() Permutation([1, 0, 3, 2]) >>> _.rank() 7 >>> p.unrank_lex(p.size, rank=7) Permutation([1, 0, 3, 2])
The product of two permutations p and q is defined as their composition as functions, (p*q)(i) = q(p(i)) [R84].
>>> p = Permutation([1, 0, 2, 3]) >>> q = Permutation([2, 3, 1, 0]) >>> list(q*p) [2, 3, 0, 1] >>> list(p*q) [3, 2, 1, 0] >>> [q(p(i)) for i in range(p.size)] [3, 2, 1, 0]
The permutation can be ‘applied’ to any list-like object, not only Permutations:
>>> p(['zero', 'one', 'four', 'two']) ['one', 'zero', 'four', 'two'] >>> p('zo42') ['o', 'z', '4', '2']
If you have a list of arbitrary elements, the corresponding permutation can be found with the from_sequence method:
>>> Permutation.from_sequence('SymPy') Permutation([1, 3, 2, 0, 4])
Checking If A Permutation Is Contained In A Group
Generally if you have a group of permutations G on n symbols, and you’re checking if a permutation on less than n symbols is part of that group, the check will fail.
Here is an example for n=5 and we check if the cycle (1,2,3) is in G:
>>> from sympy import init_printing >>> init_printing(perm_cyclic=True, pretty_print=False) >>> from sympy.combinatorics import Cycle, Permutation >>> from sympy.combinatorics.perm_groups import PermutationGroup >>> G = PermutationGroup(Cycle(2, 3)(4, 5), Cycle(1, 2, 3, 4, 5)) >>> p1 = Permutation(Cycle(2, 5, 3)) >>> p2 = Permutation(Cycle(1, 2, 3)) >>> a1 = Permutation(Cycle(1, 2, 3).list(6)) >>> a2 = Permutation(Cycle(1, 2, 3)(5)) >>> a3 = Permutation(Cycle(1, 2, 3),size=6) >>> for p in [p1,p2,a1,a2,a3]: p, G.contains(p) ((2 5 3), True) ((1 2 3), False) ((5)(1 2 3), True) ((5)(1 2 3), True) ((5)(1 2 3), True)
The check for p2 above will fail.
Checking if p1 is in G works because SymPy knows G is a group on 5 symbols, and p1 is also on 5 symbols (its largest element is 5).
For
a1
, the.list(6)
call will extend the permutation to 5 symbols, so the test will work as well. In the case ofa2
the permutation is being extended to 5 symbols by using a singleton, and in the case ofa3
it’s extended through the constructor argumentsize=6
.There is another way to do this, which is to tell the
contains
method that the number of symbols the group is on does not need to match perfectly the number of symbols for the permutation:>>> G.contains(p2,strict=False) True
This can be via the
strict
argument to thecontains
method, and SymPy will try to extend the permutation on its own and then perform the containment check.See also
References
[R79]Skiena, S. ‘Permutations.’ 1.1 in Implementing Discrete Mathematics Combinatorics and Graph Theory with Mathematica. Reading, MA: Addison-Wesley, pp. 3-16, 1990.
[R80] (1,2)Knuth, D. E. The Art of Computer Programming, Vol. 4: Combinatorial Algorithms, 1st ed. Reading, MA: Addison-Wesley, 2011.
[R81]Wendy Myrvold and Frank Ruskey. 2001. Ranking and unranking permutations in linear time. Inf. Process. Lett. 79, 6 (September 2001), 281-284. DOI=10.1016/S0020-0190(01)00141-7
[R82]D. L. Kreher, D. R. Stinson ‘Combinatorial Algorithms’ CRC Press, 1999
[R83]Graham, R. L.; Knuth, D. E.; and Patashnik, O. Concrete Mathematics: A Foundation for Computer Science, 2nd ed. Reading, MA: Addison-Wesley, 1994.
- apply(i)[source]#
Apply the permutation to an expression.
- Parameters:
i : Expr
It should be an integer between \(0\) and \(n-1\) where \(n\) is the size of the permutation.
If it is a symbol or a symbolic expression that can have integer values, an
AppliedPermutation
object will be returned which can represent an unevaluated function.
Notes
Any permutation can be defined as a bijective function \(\sigma : \{ 0, 1, \dots, n-1 \} \rightarrow \{ 0, 1, \dots, n-1 \}\) where \(n\) denotes the size of the permutation.
The definition may even be extended for any set with distinctive elements, such that the permutation can even be applied for real numbers or such, however, it is not implemented for now for computational reasons and the integrity with the group theory module.
This function is similar to the
__call__
magic, however,__call__
magic already has some other applications like permuting an array or attaching new cycles, which would not always be mathematically consistent.This also guarantees that the return type is a SymPy integer, which guarantees the safety to use assumptions.
- property array_form#
Return a copy of the attribute _array_form Examples ========
>>> from sympy.combinatorics import Permutation >>> p = Permutation([[2, 0], [3, 1]]) >>> p.array_form [2, 3, 0, 1] >>> Permutation([[2, 0, 3, 1]]).array_form [3, 2, 0, 1] >>> Permutation([2, 0, 3, 1]).array_form [2, 0, 3, 1] >>> Permutation([[1, 2], [4, 5]]).array_form [0, 2, 1, 3, 5, 4]
- ascents()[source]#
Returns the positions of ascents in a permutation, ie, the location where p[i] < p[i+1]
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([4, 0, 1, 3, 2]) >>> p.ascents() [1, 2]
See also
descents
,inversions
,min
,max
- atoms()[source]#
Returns all the elements of a permutation
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation([0, 1, 2, 3, 4, 5]).atoms() {0, 1, 2, 3, 4, 5} >>> Permutation([[0, 1], [2, 3], [4, 5]]).atoms() {0, 1, 2, 3, 4, 5}
- property cardinality#
Returns the number of all possible permutations.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.cardinality 24
- commutator(x)[source]#
Return the commutator of
self
andx
:~x*~self*x*self
If f and g are part of a group, G, then the commutator of f and g is the group identity iff f and g commute, i.e. fg == gf.
Examples
>>> from sympy.combinatorics import Permutation >>> from sympy import init_printing >>> init_printing(perm_cyclic=False, pretty_print=False) >>> p = Permutation([0, 2, 3, 1]) >>> x = Permutation([2, 0, 3, 1]) >>> c = p.commutator(x); c Permutation([2, 1, 3, 0]) >>> c == ~x*~p*x*p True
>>> I = Permutation(3) >>> p = [I + i for i in range(6)] >>> for i in range(len(p)): ... for j in range(len(p)): ... c = p[i].commutator(p[j]) ... if p[i]*p[j] == p[j]*p[i]: ... assert c == I ... else: ... assert c != I ...
References
- commutes_with(other)[source]#
Checks if the elements are commuting.
Examples
>>> from sympy.combinatorics import Permutation >>> a = Permutation([1, 4, 3, 0, 2, 5]) >>> b = Permutation([0, 1, 2, 3, 4, 5]) >>> a.commutes_with(b) True >>> b = Permutation([2, 3, 5, 4, 1, 0]) >>> a.commutes_with(b) False
- property cycle_structure#
Return the cycle structure of the permutation as a dictionary indicating the multiplicity of each cycle length.
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation(3).cycle_structure {1: 4} >>> Permutation(0, 4, 3)(1, 2)(5, 6).cycle_structure {2: 2, 3: 1}
- property cycles#
Returns the number of cycles contained in the permutation (including singletons).
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation([0, 1, 2]).cycles 3 >>> Permutation([0, 1, 2]).full_cyclic_form [[0], [1], [2]] >>> Permutation(0, 1)(2, 3).cycles 2
- property cyclic_form#
This is used to convert to the cyclic notation from the canonical notation. Singletons are omitted.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([0, 3, 1, 2]) >>> p.cyclic_form [[1, 3, 2]] >>> Permutation([1, 0, 2, 4, 3, 5]).cyclic_form [[0, 1], [3, 4]]
See also
- descents()[source]#
Returns the positions of descents in a permutation, ie, the location where p[i] > p[i+1]
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([4, 0, 1, 3, 2]) >>> p.descents() [0, 3]
See also
ascents
,inversions
,min
,max
- classmethod from_inversion_vector(inversion)[source]#
Calculates the permutation from the inversion vector.
Examples
>>> from sympy.combinatorics import Permutation >>> from sympy import init_printing >>> init_printing(perm_cyclic=False, pretty_print=False) >>> Permutation.from_inversion_vector([3, 2, 1, 0, 0]) Permutation([3, 2, 1, 0, 4, 5])
- classmethod from_sequence(i, key=None)[source]#
Return the permutation needed to obtain
i
from the sorted elements ofi
. If custom sorting is desired, a key can be given.Examples
>>> from sympy.combinatorics import Permutation
>>> Permutation.from_sequence('SymPy') (4)(0 1 3) >>> _(sorted("SymPy")) ['S', 'y', 'm', 'P', 'y'] >>> Permutation.from_sequence('SymPy', key=lambda x: x.lower()) (4)(0 2)(1 3)
- property full_cyclic_form#
Return permutation in cyclic form including singletons.
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation([0, 2, 1]).full_cyclic_form [[0], [1, 2]]
- get_adjacency_distance(other)[source]#
Computes the adjacency distance between two permutations.
Explanation
This metric counts the number of times a pair i,j of jobs is adjacent in both p and p’. If n_adj is this quantity then the adjacency distance is n - n_adj - 1 [1]
[1] Reeves, Colin R. Landscapes, Operators and Heuristic search, Annals of Operational Research, 86, pp 473-490. (1999)
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([0, 3, 1, 2, 4]) >>> q = Permutation.josephus(4, 5, 2) >>> p.get_adjacency_distance(q) 3 >>> r = Permutation([0, 2, 1, 4, 3]) >>> p.get_adjacency_distance(r) 4
- get_adjacency_matrix()[source]#
Computes the adjacency matrix of a permutation.
Explanation
If job i is adjacent to job j in a permutation p then we set m[i, j] = 1 where m is the adjacency matrix of p.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation.josephus(3, 6, 1) >>> p.get_adjacency_matrix() Matrix([ [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 0, 1], [0, 1, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0]]) >>> q = Permutation([0, 1, 2, 3]) >>> q.get_adjacency_matrix() Matrix([ [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1], [0, 0, 0, 0]])
- get_positional_distance(other)[source]#
Computes the positional distance between two permutations.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([0, 3, 1, 2, 4]) >>> q = Permutation.josephus(4, 5, 2) >>> r = Permutation([3, 1, 4, 0, 2]) >>> p.get_positional_distance(q) 12 >>> p.get_positional_distance(r) 12
- get_precedence_distance(other)[source]#
Computes the precedence distance between two permutations.
Explanation
Suppose p and p’ represent n jobs. The precedence metric counts the number of times a job j is preceded by job i in both p and p’. This metric is commutative.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([2, 0, 4, 3, 1]) >>> q = Permutation([3, 1, 2, 4, 0]) >>> p.get_precedence_distance(q) 7 >>> q.get_precedence_distance(p) 7
- get_precedence_matrix()[source]#
Gets the precedence matrix. This is used for computing the distance between two permutations.
Examples
>>> from sympy.combinatorics import Permutation >>> from sympy import init_printing >>> init_printing(perm_cyclic=False, pretty_print=False) >>> p = Permutation.josephus(3, 6, 1) >>> p Permutation([2, 5, 3, 1, 4, 0]) >>> p.get_precedence_matrix() Matrix([ [0, 0, 0, 0, 0, 0], [1, 0, 0, 0, 1, 0], [1, 1, 0, 1, 1, 1], [1, 1, 0, 0, 1, 0], [1, 0, 0, 0, 0, 0], [1, 1, 0, 1, 1, 0]])
- index()[source]#
Returns the index of a permutation.
The index of a permutation is the sum of all subscripts j such that p[j] is greater than p[j+1].
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([3, 0, 2, 1, 4]) >>> p.index() 2
- inversion_vector()[source]#
Return the inversion vector of the permutation.
The inversion vector consists of elements whose value indicates the number of elements in the permutation that are lesser than it and lie on its right hand side.
The inversion vector is the same as the Lehmer encoding of a permutation.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([4, 8, 0, 7, 1, 5, 3, 6, 2]) >>> p.inversion_vector() [4, 7, 0, 5, 0, 2, 1, 1] >>> p = Permutation([3, 2, 1, 0]) >>> p.inversion_vector() [3, 2, 1]
The inversion vector increases lexicographically with the rank of the permutation, the -ith element cycling through 0..i.
>>> p = Permutation(2) >>> while p: ... print('%s %s %s' % (p, p.inversion_vector(), p.rank())) ... p = p.next_lex() (2) [0, 0] 0 (1 2) [0, 1] 1 (2)(0 1) [1, 0] 2 (0 1 2) [1, 1] 3 (0 2 1) [2, 0] 4 (0 2) [2, 1] 5
See also
- inversions()[source]#
Computes the number of inversions of a permutation.
Explanation
An inversion is where i > j but p[i] < p[j].
For small length of p, it iterates over all i and j values and calculates the number of inversions. For large length of p, it uses a variation of merge sort to calculate the number of inversions.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([0, 1, 2, 3, 4, 5]) >>> p.inversions() 0 >>> Permutation([3, 2, 1, 0]).inversions() 6
References
- property is_Empty#
Checks to see if the permutation is a set with zero elements
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation([]).is_Empty True >>> Permutation([0]).is_Empty False
See also
- property is_Identity#
Returns True if the Permutation is an identity permutation.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([]) >>> p.is_Identity True >>> p = Permutation([[0], [1], [2]]) >>> p.is_Identity True >>> p = Permutation([0, 1, 2]) >>> p.is_Identity True >>> p = Permutation([0, 2, 1]) >>> p.is_Identity False
See also
- property is_Singleton#
Checks to see if the permutation contains only one number and is thus the only possible permutation of this set of numbers
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation([0]).is_Singleton True >>> Permutation([0, 1]).is_Singleton False
See also
- property is_even#
Checks if a permutation is even.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.is_even True >>> p = Permutation([3, 2, 1, 0]) >>> p.is_even True
See also
- property is_odd#
Checks if a permutation is odd.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.is_odd False >>> p = Permutation([3, 2, 0, 1]) >>> p.is_odd True
See also
- classmethod josephus(m, n, s=1)[source]#
Return as a permutation the shuffling of range(n) using the Josephus scheme in which every m-th item is selected until all have been chosen. The returned permutation has elements listed by the order in which they were selected.
The parameter
s
stops the selection process when there ares
items remaining and these are selected by continuing the selection, counting by 1 rather than bym
.Consider selecting every 3rd item from 6 until only 2 remain:
choices chosen ======== ====== 012345 01 345 2 01 34 25 01 4 253 0 4 2531 0 25314 253140
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation.josephus(3, 6, 2).array_form [2, 5, 3, 1, 4, 0]
References
- length()[source]#
Returns the number of integers moved by a permutation.
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation([0, 3, 2, 1]).length() 2 >>> Permutation([[0, 1], [2, 3]]).length() 4
- list(size=None)[source]#
Return the permutation as an explicit list, possibly trimming unmoved elements if size is less than the maximum element in the permutation; if this is desired, setting
size=-1
will guarantee such trimming.Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation(2, 3)(4, 5) >>> p.list() [0, 1, 3, 2, 5, 4] >>> p.list(10) [0, 1, 3, 2, 5, 4, 6, 7, 8, 9]
Passing a length too small will trim trailing, unchanged elements in the permutation:
>>> Permutation(2, 4)(1, 2, 4).list(-1) [0, 2, 1] >>> Permutation(3).list(-1) []
- max()[source]#
The maximum element moved by the permutation.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([1, 0, 2, 3, 4]) >>> p.max() 1
See also
- min()[source]#
The minimum element moved by the permutation.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([0, 1, 4, 3, 2]) >>> p.min() 2
See also
- next_lex()[source]#
Returns the next permutation in lexicographical order. If self is the last permutation in lexicographical order it returns None. See [4] section 2.4.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([2, 3, 1, 0]) >>> p = Permutation([2, 3, 1, 0]); p.rank() 17 >>> p = p.next_lex(); p.rank() 18
See also
- next_nonlex()[source]#
Returns the next permutation in nonlex order [3]. If self is the last permutation in this order it returns None.
Examples
>>> from sympy.combinatorics import Permutation >>> from sympy import init_printing >>> init_printing(perm_cyclic=False, pretty_print=False) >>> p = Permutation([2, 0, 3, 1]); p.rank_nonlex() 5 >>> p = p.next_nonlex(); p Permutation([3, 0, 1, 2]) >>> p.rank_nonlex() 6
See also
- next_trotterjohnson()[source]#
Returns the next permutation in Trotter-Johnson order. If self is the last permutation it returns None. See [4] section 2.4. If it is desired to generate all such permutations, they can be generated in order more quickly with the
generate_bell
function.Examples
>>> from sympy.combinatorics import Permutation >>> from sympy import init_printing >>> init_printing(perm_cyclic=False, pretty_print=False) >>> p = Permutation([3, 0, 2, 1]) >>> p.rank_trotterjohnson() 4 >>> p = p.next_trotterjohnson(); p Permutation([0, 3, 2, 1]) >>> p.rank_trotterjohnson() 5
- order()[source]#
Computes the order of a permutation.
When the permutation is raised to the power of its order it equals the identity permutation.
Examples
>>> from sympy.combinatorics import Permutation >>> from sympy import init_printing >>> init_printing(perm_cyclic=False, pretty_print=False) >>> p = Permutation([3, 1, 5, 2, 4, 0]) >>> p.order() 4 >>> (p**(p.order())) Permutation([], size=6)
See also
- parity()[source]#
Computes the parity of a permutation.
Explanation
The parity of a permutation reflects the parity of the number of inversions in the permutation, i.e., the number of pairs of x and y such that
x > y
butp[x] < p[y]
.Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.parity() 0 >>> p = Permutation([3, 2, 0, 1]) >>> p.parity() 1
See also
- classmethod random(n)[source]#
Generates a random permutation of length
n
.Uses the underlying Python pseudo-random number generator.
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation.random(2) in (Permutation([1, 0]), Permutation([0, 1])) True
- rank()[source]#
Returns the lexicographic rank of the permutation.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.rank() 0 >>> p = Permutation([3, 2, 1, 0]) >>> p.rank() 23
See also
- rank_nonlex(inv_perm=None)[source]#
This is a linear time ranking algorithm that does not enforce lexicographic order [3].
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.rank_nonlex() 23
See also
- rank_trotterjohnson()[source]#
Returns the Trotter Johnson rank, which we get from the minimal change algorithm. See [4] section 2.4.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.rank_trotterjohnson() 0 >>> p = Permutation([0, 2, 1, 3]) >>> p.rank_trotterjohnson() 7
See also
- resize(n)[source]#
Resize the permutation to the new size
n
.- Parameters:
n : int
The new size of the permutation.
- Raises:
ValueError
If the permutation cannot be resized to the given size. This may only happen when resized to a smaller size than the original.
Examples
>>> from sympy.combinatorics import Permutation
Increasing the size of a permutation:
>>> p = Permutation(0, 1, 2) >>> p = p.resize(5) >>> p (4)(0 1 2)
Decreasing the size of the permutation:
>>> p = p.resize(4) >>> p (3)(0 1 2)
If resizing to the specific size breaks the cycles:
>>> p.resize(2) Traceback (most recent call last): ... ValueError: The permutation cannot be resized to 2 because the cycle (0, 1, 2) may break.
- static rmul(*args)[source]#
Return product of Permutations [a, b, c, …] as the Permutation whose ith value is a(b(c(i))).
a, b, c, … can be Permutation objects or tuples.
Examples
>>> from sympy.combinatorics import Permutation
>>> a, b = [1, 0, 2], [0, 2, 1] >>> a = Permutation(a); b = Permutation(b) >>> list(Permutation.rmul(a, b)) [1, 2, 0] >>> [a(b(i)) for i in range(3)] [1, 2, 0]
This handles the operands in reverse order compared to the
*
operator:>>> a = Permutation(a); b = Permutation(b) >>> list(a*b) [2, 0, 1] >>> [b(a(i)) for i in range(3)] [2, 0, 1]
Notes
All items in the sequence will be parsed by Permutation as necessary as long as the first item is a Permutation:
>>> Permutation.rmul(a, [0, 2, 1]) == Permutation.rmul(a, b) True
The reverse order of arguments will raise a TypeError.
- classmethod rmul_with_af(*args)[source]#
same as rmul, but the elements of args are Permutation objects which have _array_form
- runs()[source]#
Returns the runs of a permutation.
An ascending sequence in a permutation is called a run [5].
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([2, 5, 7, 3, 6, 0, 1, 4, 8]) >>> p.runs() [[2, 5, 7], [3, 6], [0, 1, 4, 8]] >>> q = Permutation([1,3,2,0]) >>> q.runs() [[1, 3], [2], [0]]
- signature()[source]#
Gives the signature of the permutation needed to place the elements of the permutation in canonical order.
The signature is calculated as (-1)^<number of inversions>
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([0, 1, 2]) >>> p.inversions() 0 >>> p.signature() 1 >>> q = Permutation([0,2,1]) >>> q.inversions() 1 >>> q.signature() -1
See also
- property size#
Returns the number of elements in the permutation.
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation([[3, 2], [0, 1]]).size 4
See also
- support()[source]#
Return the elements in permutation, P, for which P[i] != i.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([[3, 2], [0, 1], [4]]) >>> p.array_form [1, 0, 3, 2, 4] >>> p.support() [0, 1, 2, 3]
- transpositions()[source]#
Return the permutation decomposed into a list of transpositions.
Explanation
It is always possible to express a permutation as the product of transpositions, see [1]
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([[1, 2, 3], [0, 4, 5, 6, 7]]) >>> t = p.transpositions() >>> t [(0, 7), (0, 6), (0, 5), (0, 4), (1, 3), (1, 2)] >>> print(''.join(str(c) for c in t)) (0, 7)(0, 6)(0, 5)(0, 4)(1, 3)(1, 2) >>> Permutation.rmul(*[Permutation([ti], size=p.size) for ti in t]) == p True
References
- classmethod unrank_lex(size, rank)[source]#
Lexicographic permutation unranking.
Examples
>>> from sympy.combinatorics import Permutation >>> from sympy import init_printing >>> init_printing(perm_cyclic=False, pretty_print=False) >>> a = Permutation.unrank_lex(5, 10) >>> a.rank() 10 >>> a Permutation([0, 2, 4, 1, 3])
- classmethod unrank_nonlex(n, r)[source]#
This is a linear time unranking algorithm that does not respect lexicographic order [3].
Examples
>>> from sympy.combinatorics import Permutation >>> from sympy import init_printing >>> init_printing(perm_cyclic=False, pretty_print=False) >>> Permutation.unrank_nonlex(4, 5) Permutation([2, 0, 3, 1]) >>> Permutation.unrank_nonlex(4, -1) Permutation([0, 1, 2, 3])
See also
- classmethod unrank_trotterjohnson(size, rank)[source]#
Trotter Johnson permutation unranking. See [4] section 2.4.
Examples
>>> from sympy.combinatorics import Permutation >>> from sympy import init_printing >>> init_printing(perm_cyclic=False, pretty_print=False) >>> Permutation.unrank_trotterjohnson(5, 10) Permutation([0, 3, 1, 2, 4])
See also
- class sympy.combinatorics.permutations.Cycle(*args)[source]#
Wrapper around dict which provides the functionality of a disjoint cycle.
Explanation
A cycle shows the rule to use to move subsets of elements to obtain a permutation. The Cycle class is more flexible than Permutation in that 1) all elements need not be present in order to investigate how multiple cycles act in sequence and 2) it can contain singletons:
>>> from sympy.combinatorics.permutations import Perm, Cycle
A Cycle will automatically parse a cycle given as a tuple on the rhs:
>>> Cycle(1, 2)(2, 3) (1 3 2)
The identity cycle, Cycle(), can be used to start a product:
>>> Cycle()(1, 2)(2, 3) (1 3 2)
The array form of a Cycle can be obtained by calling the list method (or passing it to the list function) and all elements from 0 will be shown:
>>> a = Cycle(1, 2) >>> a.list() [0, 2, 1] >>> list(a) [0, 2, 1]
If a larger (or smaller) range is desired use the list method and provide the desired size – but the Cycle cannot be truncated to a size smaller than the largest element that is out of place:
>>> b = Cycle(2, 4)(1, 2)(3, 1, 4)(1, 3) >>> b.list() [0, 2, 1, 3, 4] >>> b.list(b.size + 1) [0, 2, 1, 3, 4, 5] >>> b.list(-1) [0, 2, 1]
Singletons are not shown when printing with one exception: the largest element is always shown – as a singleton if necessary:
>>> Cycle(1, 4, 10)(4, 5) (1 5 4 10) >>> Cycle(1, 2)(4)(5)(10) (1 2)(10)
The array form can be used to instantiate a Permutation so other properties of the permutation can be investigated:
>>> Perm(Cycle(1, 2)(3, 4).list()).transpositions() [(1, 2), (3, 4)]
Notes
The underlying structure of the Cycle is a dictionary and although the __iter__ method has been redefined to give the array form of the cycle, the underlying dictionary items are still available with the such methods as items():
>>> list(Cycle(1, 2).items()) [(1, 2), (2, 1)]
See also
- list(size=None)[source]#
Return the cycles as an explicit list starting from 0 up to the greater of the largest value in the cycles and size.
Truncation of trailing unmoved items will occur when size is less than the maximum element in the cycle; if this is desired, setting
size=-1
will guarantee such trimming.Examples
>>> from sympy.combinatorics import Cycle >>> p = Cycle(2, 3)(4, 5) >>> p.list() [0, 1, 3, 2, 5, 4] >>> p.list(10) [0, 1, 3, 2, 5, 4, 6, 7, 8, 9]
Passing a length too small will trim trailing, unchanged elements in the permutation:
>>> Cycle(2, 4)(1, 2, 4).list(-1) [0, 2, 1]
- sympy.combinatorics.permutations._af_parity(pi)[source]#
Computes the parity of a permutation in array form.
Explanation
The parity of a permutation reflects the parity of the number of inversions in the permutation, i.e., the number of pairs of x and y such that x > y but p[x] < p[y].
Examples
>>> from sympy.combinatorics.permutations import _af_parity >>> _af_parity([0, 1, 2, 3]) 0 >>> _af_parity([3, 2, 0, 1]) 1
See also
Generators#
- generators.symmetric()[source]#
Generates the symmetric group of order n, Sn.
Examples
>>> from sympy.combinatorics.generators import symmetric >>> list(symmetric(3)) [(2), (1 2), (2)(0 1), (0 1 2), (0 2 1), (0 2)]
- generators.cyclic()[source]#
Generates the cyclic group of order n, Cn.
Examples
>>> from sympy.combinatorics.generators import cyclic >>> list(cyclic(5)) [(4), (0 1 2 3 4), (0 2 4 1 3), (0 3 1 4 2), (0 4 3 2 1)]
See also
- generators.alternating()[source]#
Generates the alternating group of order n, An.
Examples
>>> from sympy.combinatorics.generators import alternating >>> list(alternating(3)) [(2), (0 1 2), (0 2 1)]
- generators.dihedral()[source]#
Generates the dihedral group of order 2n, Dn.
The result is given as a subgroup of Sn, except for the special cases n=1 (the group S2) and n=2 (the Klein 4-group) where that’s not possible and embeddings in S2 and S4 respectively are given.
Examples
>>> from sympy.combinatorics.generators import dihedral >>> list(dihedral(3)) [(2), (0 2), (0 1 2), (1 2), (0 2 1), (2)(0 1)]
See also